Solve s^3+6s^2+11s+6=0

$$s ^ { 3 } + 6 s ^ { 2 } + 11 s + 6 = 0$$

Solve for s

$$s=-2$$

$$s=-1$$

$$s=-3$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 6 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±6,±3,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$s=-1$$

By Factor theorem, s-k is a factor of the polynomial for each root k. Divide s^{3}+6s^{2}+11s+6 by s+1 to get s^{2}+5s+6. Solve the equation where the result equals to 0.

$$s^{2}+5s+6=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 5 for b, and 6 for c in the quadratic formula.

$$s=\frac{-5±\sqrt{5^{2}-4\times 1\times 6}}{2}$$

Do the calculations.

$$s=\frac{-5±1}{2}$$

Solve the equation s^{2}+5s+6=0 when ± is plus and when ± is minus.

$$s=-3 s=-2$$

List all found solutions.

$$s=-1 s=-3 s=-2$$